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4 years ago

Solve the inequality: |6x + 7| ≥ −3 and write the solution in interval notation

1 answer:

8 0

Hello :
 |6x + 7| ≥ −3 is right (true) for all reals : x because |6x + 7| ≥ 0
the set solution is : R
in interval notation : ]-∞,+∞[

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Add the equations to find the solutions to the system.

amm1812

The answer is C, (1,-4) the y automatically cancels out and makes 10x=10. X=1 and x is always first so it automatically makes C the answer.

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3 years ago

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I can't stop NOT understanding this problem.. 100 - 1 = ?

Butoxors [25]

Answer: the answer is 99

Step-by-step explanation: this is esay

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2 years ago

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A bag contains 4 red marbles and 1 yellow marble. What is the probability of drawing a red marble, then replacing that marble, t

ki77a [65]

Answer:

16/25

Step-by-step explanation:

We have 4 red marbles and a total of 4 + 1 = 5 marbles.

Probability is simply (# times a specific event can occur) / (# times any event can occur).

Here, the probability of drawing a red marble is: 4/5 (because there are 4 red marbles to choose from and a total of 5 marbles to choose from).

We replace the marble afterward, so the probability that the second pick is also red would still be 4/5.

Then, multiply the two fractions together: (4/5) * (4/5) = 16/25.

Hope this helps!

8 0

3 years ago

In the △ABC, the height AN = 24 in, BN = 18 in, AC = 40 in. Find AB and BC. Consider all possible cases. Case 1 : N∈BC, AB = __,

aivan3 [116]

Answer:

Case 1:

$AB = 30$

$BC = 50$

Case 2:

$AB = 15.9$

$BC = 36.7$

Case 3: Not possible

Step-by-step explanation:

Given

See attachment for illustration of each case

Required

Find AB and BC

Case 1:

Using Pythagoras theorem in ANB, we have:

$AB^2 = AN^2 + BN^2$

This gives:

$AB^2 = 24^2 + 18^2$

$AB^2 = 576 + 324$

$AB^2 = 900$

Take square roots of both sides

$AB = \sqrt{900$

$AB = 30$

To calculate BC, we consider ANC, where:

$AC^2 = AN^2 + NC^2$

$40^2 = 24^2 + NC^2$

$1600 = 576 + NC^2$

Collect like terms

$NC^2 = 1600 - 576$

$NC^2 = 1024$

Take square roots

$NC = \sqrt{1024$

$NC = 32$

So:

$BC = NC + BN$

$BC = 32 + 18$

$BC = 50$

Case 2:

Using Pythagoras theorem in ANB, we have:

$AN^2 = AB^2 + BN^2$

This gives:

$24^2 = AB^2 + 18^2$

$576 = AB^2 + 324$

Collect like terms

$AB^2 = 576 - 324$

$AB^2 = 252$

Take square roots of both sides

$AB = \sqrt{252$

$AB = 15.9$

To calculate BC, we consider ABC, where:

$AC^2 = AB^2 + BC^2$

$40^2 = 252 + BC^2$

$1600 = 252 + BC^2$

Collect like terms

$BC^2 = 1600 - 252$

$BC^2 = 1348$

Take square roots

$BC = \sqrt{1348$

$BC = 36.7$

Case 3:

This is not possible because in ANC

The hypotenuse AN (24) is less than AC (40)

3 0

3 years ago

True or false? The volume of a sphere is two-thirds the volume of a cylinder with the same radius r and height 2r. Explain your

Margaret [11]

You can test this like this:
by dividing the formula of the sphere on the formula of the cylinder
((4/3)*PI*r^3)/((2r*r^2*PI)=
((4/3)*r^3)/((1/3)*2r^3)=

multiply by 3 both above and under to get rid of the fraction
4*r^3/2r^3*3
4/2*3
4/6=2/3

This should work as reasoning :)

8 0

3 years ago